Question

已知tana=-1/2,求1+2sinacosa/sin²a-cos²a的值是

A.1/3 B.3 C.-1/3 D.-3
有过程最好，3Q

Answer 1

1+2sinacosa/sin²a-cos²a
=sin²a+cos²a+2sinacosa/sin²a-cos²a
=tan²a+1+2tana/(tan²a-1)
=[1/4+1-1]/(1/4-1)
=(1/4)/(-3/4)
=-1/3
选C

Answer 2

1+2sinacosa/sin²a-cos²a
=(sin^2a+cos^2+2sinacosa)/(sin^2a-cos^2)
=(tan^2a+1+2tana)/(tan^2-1)
=(tana+1)^2/(tan^2-1)
=(tana+1)/(tana-1)
=(1/2)/(-3/2)
=-1/3
C

Answer 3

(1+2sinαcosα)/(sin²α-cos²α)=(sin²α+cos²α+2sinαcosα)/(sin²α-cos²α)
=(sinα+cosα)²/[(sinα+cosα)(sinα-cosα)]
=(sinα+cosα)/(sinα-cosα)
=(tanα+1)/(tanα-1) (上下同除以cosα)
= -1/3.

Answer 4

原式为1-tan2A,再用正切二倍角公式得-1/3